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Clicker Question
Room Frequency BA
A solid piece of plastic of volume V, and density ρplastic is floating
partially submerged in a cup of water. (The density of water is ρwater.)
What is the buoyant force on the plastic?
A) Zero
B) ρplastic V
C) ρwater V
D) ρwater V g
E) ρplastic V g
FB
mplasticg
The plastic is in equilibrium so FB = mplasticg = ρplastic V g !
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Announcements
• CAPA assignment #13 is due on Friday at 10 pm.
• This week in Section: Assignment #6
• Start reading Chapter 11 on Vibrations and Waves
• I will have regular office hours 1:45 – 3:45 in the
Physics Helproom today
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Fluids in Motion: Fluid Dynamics
Many, many different types of motion depending on
particular properties of fluid: waves, rivers, geysers,
tornados, hurricanes, ocean currents, trade winds,
whirlpools, eddies, tsunamis, earthquakes, and on and on!
We’ll focus on the simplest motion: flow
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Fluids in Motion: Flow
Two main types of flow: Laminar and Turbulent
Laminar
Turbulent
We’ll focus on the simplest flow: laminar flow
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Analysis of Flow
Analyzing flow at the force level is mathematically complex 
Use conservation laws!
1) Conservation of Mass: the Continuity Equation
2) Conservation of Energy: Bernoulli’s Equation
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Continuity Equation
Consider the flow of a fluid through a pipe in which the cross sectional
area changes from A1 to A2
The mass of fluid going in has to equal the mass of fluid coming out:
conservation of mass!
The speed of the fluids must be different!
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Continuity Equation
To analyze this mass conservation, we calculate the mass flow rate: m
mass flowing through a volume

elapsed time
m1
Flow rate in =
t
must equal
m2
Flow rate out =
t
t
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Mass Flow Rates
Flow rate in = m1  1V1
t
t

1 A1l1
t
 1 A1v1
m2
 2 A2 v2
Flow rate out =
t
Continuity Equation: 1 A1v1  2 A2 v2
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Continuity for Incompressible Fluids
If the fluid is incompressible: ρ1 = ρ2 so
A1v1  A2 v2
Does this make sense?
A1
v2 
v1
A2
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Room Frequency BA
Clicker Question
“Incompressible” blood flows out of the heart via the aorta at a
speed vaorta. The radius of the aorta raorta = 1.2 cm. What is the speed
of the blood in a connecting artery whose radius is 0.6 cm?
A)
B)
C)
D)
E)
vaorta
2 vaorta
(2)1/2 vaorta
4 vaorta
8 vaorta
vartery
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 Aaorta 
  raorta


vaorta   2  vaorta

 Aartery 
  rartery 
2
 raorta 

vaorta  4vaorta

 rartery 
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Bernoulli’s Equation: Conservation of Energy
Earlier in the course we learned: PEi  KEi  Wi  PE f  KE f  W f
Applied to fluid flow, we consider energy of pieces of fluid of mass Δm
m1gy1  12 m1v12  (P1A1l1 )  m2 gy2  12 m2 v22  (P2 A2 l2 )
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Bernoulli’s Equation: Incompressible Fluids
Now using m1  V1   A1l1 and m2  V1   A2 l2
and the continuity equation A1v1  A2 v2 you get
Bernoulli’s Equation: gy1  12 v12  P1  gy2  12 v22  P2
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Applications of Bernoulli’s Equation
Static Fluid (v=0):  gy1  P1   gy2  P2
y1  y2 case:
1
2
v12  P1  12 v22  P2
Bernoulli’s Equation is behind many common phenomena!
1)
2)
3)
4)
5)
6)
7)
8)
Curve balls
Aerodynamic Lift
Sailing into the wind
Transient Ischemic Attacks (“mini-strokes”)
Light objects getting sucked out your car window
Shower curtains bowing in
Flat roofs flying off houses in Boulder!
Ping pong ball demo
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Flat Roof Example
Wind flows over a flat roof with area A = 240 m2 at a speed of voutside
= 35 m/s (125 km/h = 80 mi/h). What net force does the wind apply
to the roof?
Inside
Outside
1 2
1 2
gy1  v1  P1  gy2  v2  P2
2
2
1
1 2
2
 gh   vinside  Pinside   gh  voutside  Poutside
2
2
1
F
2
Pinside  Poutside   voutside 
2
A
1
2
F   Avoutside
 0.5(1.29kg / m 3 )(240m 2 )(35m / s)2
2
 1lb 
5
 1.89x10 N 
 42, 600lbs

 4.45N 

v = 35 m/s
hh h

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Room Frequency BA
Clicker Question
F = lift=(Pbot-Ptop)(Wing Area)
For an airplane wing (an
air-foil) the upward lift
force is derivable from
Bernoulli’s equation. How
does the air speed over
the wing compare to the
air speed under the wing?
It is……
A)
B)
C)
D)
Faster
Slower
Same
Unknown
faster
1
2
1
2
slower
2
2
 gytop  vtop
 Ptop   gybot  vbot
 Pbot
1 2
1 2
vtop  Ptop  vbot
 Pbot
2
2
On the top side, the air has to travel farther to meet at the back
edge of the wing!
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Oscillations!
Throughout nature things are bound together by forces
which allow things to oscillate back and forth.
It is important to get a deeper understanding of these phenomena!
We’ll focus on the most common and the most simple
oscillation: Simple Harmonic Motion (SHM)
Requirements for SHM:
1) There is a restoring force proportional to the displacement from
equilibrium
2) The range of the motion (amplitude) is independent of the
frequency
3) The position, velocity, and acceleration are all sinusoidal
(harmonic) in time
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Mass and Spring
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A Simple Harmonic Oscillator: Spring and Mass!
F  ma  kx
(Hooke's Law)
Note:
• restoring force is proportional to displacement
• force is not constant, so acceleration isn’t either: a = -(k/m)x
• “amplitude” A is the maximum displacement xmax, occurs with v = 0
• mass oscillates between x = A & x = -A
• maximum speed vmax occurs when displacement x = 0
• a “cycle” is the full extent of motion as shown
• the time to complete one cycle is the “period” T
• frequency is the number of cycles per second: f = 1/T (units Hz)
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